Let me start by posting a question, which is essentially a combinatorial question. I will post below at least a short explanation what lead me to ask this question.
Let us work with the set $\omega^\omega$ (i.e., all sequences of positive integers) with a pointwise ordering $$x\le y \Leftrightarrow (\forall n\in \omega) x_n\le y_n$$
Does there exist a system $\mathcal B\subseteq\omega^\omega$ such that
- $(\forall x\in\omega^\omega)(\exists y\in\mathcal B) x\le y$ (i.e., $\mathcal B$ is cofinal in $\omega^\omega$)
- For any $x\in\omega^\omega$ the set $\{z\in\mathcal B; z\le x\}$ is finite.
My motivation was that I was thinking about a problem posted by Michael Greinecker in this question. He asks there whether we can have a local base at $x$ such that no intersection of a countable subsystem of this base is a neighborhood of $x$; under some conditions on the point $x$ and the space $X$.
I wanted to start with some simple examples to get better understanding of the problem. I was thinking about the $X$ which is obtained from topological sum of countably many copies of Alexandroff compactification of a countable discrete space, if we glue the non-isolated point from each copy into one point.
Alexandroff compactification of a discrete space is simply a convergent sequence (it is homeomorphic to $\{0\}\cup\{1/n; n=1,2,\dots\}$ with the topology inherited from the real line.) So this space can be visualized like in the following picture. (Each ray in this picture is represents a convergent sequence and a typical basic set is marked there. Ignore the labels given there - I have simply used a picture I had prepared before for different purposes; a notation in that situation was completely different.)
Or, if you prefer another representation of this space, you can take $Y=\{n+1/k; n,k\in\mathbb N\}$ and then $X\cong Y/\mathbb N$.
This space is countably tight, since the class of countably tight spaces is closed under topological sums and quotient maps. EDIT: Alex Ravsky pointed out in a comment (now deleted), that this is unnecessary complicated - this space is countably tight simply because it is countable. Anyway, the argument I gave would work for arbitrarily many copies, too. (Or if we are using a different countably tight space instead of the convergent sequence.) So it might still be interesting for other spaces of similar type.
This space has only one non-isolated point and it is not first countable. So this spaces fulfills the assumptions from the other question. (Hence if no family with the above properties exists, this would yield a counterexample.)
Let us denote the non-isolated point of $X$ by $\infty$. For each sequence $x\in\omega^\omega$ we get a neighborhood of $\infty$ simply by taking all numbers greater than or equal to $x_n$ in the $n$-th copy of the convergent sequence. Such neighborhoods comprise a local base at $\infty$.
So the cofinality condition for the system $\mathcal B$ means that if we take the neighborhoods corresponding to sequences from $\mathcal B$, they form a local base.
The second condition correspond to the assumption that no infinite countable intersection of (distinct) sets from this local base will be again an open set.
If seems that there exists no family $\cal B$ with the above properties. Suppose the opposite. Since the family $\cal B$ is cofinal in $\omega^\omega$, the diagonal procedure shows that $\cal B$ is uncountable (moreover, maybe the minimal cardinality of $\cal B$ is related with so-called small cardinals, like $\operatorname{cof}(\omega^\omega)$).
Now by induction we can build an non-increasing sequence $\{\cal B_n:n\in \omega\}$ of uncountable subfamilies of $\cal B$ such that for each $n$, each $x,y\in\cal B_n$ and each $i\le n$ holds $x(i)=y(i)$. Again by an induction we choose a sequence $\{x_n\}$ of distinct functions such that $x_n\in\cal B_n$ for each $n$. This condition implies that for each $m$ the set $\{x_n(m):n\in\omega\}$ has a finite supremum $x(m)$. Since $\cal B$ is cofinal, there exists a sequence $y\in\cal B$ such that $y\ge x$. Then $y\ge x_n$ for each $n$, which contradicts the second property of $\cal B$.